# Geometry and tangent-chord theorem problem?

According to the figure, CA is tangent to the circle, centre O, at A. ABT and POT are straight lines.

Question:

Given that BT is equal to the radius of the circle, prove that:

1. $$\angle ABP = 3 \angle OBP$$
2. $$\angle POA = 3 \angle BOT$$
3. $$\angle OBT = 2 (\angle BTO + \angle CAB)$$ My attempt:

• $$\angle AOP = 2 \angle ABP.$$
• $$\angle OBP = \angle OPB.$$
• $$\angle CAB = \angle APO.$$
• $$\angle BOA = 2 \angle APB.$$
• What is the question? May 26 '20 at 6:58
• prove that: 1. Angle ABP = 3 time of Angle OBP 2. Angle POA = 3 times of Angle BOT 3. Angle OBT = 2 times of (Angle BTO + Angle CAB) May 26 '20 at 6:59
• Yes, I have edited the original post to include my attempts. May 26 '20 at 7:10
• Use the isosceles properties of BOT, AOB and AOP to relate the angles along with the AOP = 2* ABP. That should solve all of them May 26 '20 at 7:23

a):

$$BOT=2PBO$$

$$ABO=2BOT=4PBO$$

$$ABP=ABO-PBO=4PBO-PBI=3PBO$$

b):

$$ABP+\frac{1}{3}ABP=2BOT$$

$$\frac{4}{3}ABP=2BOT$$

$$ABP=\frac{1}{2}POA$$

$$\frac{4}{3}(\frac{1}{2}POA)=2BOT$$

Therefore:

$$POA=3BOT$$

c):

$$OBT=AOB+OBA$$

$$AOB=2ABP=2CAB$$

$$OBA=2BTO$$

Therefore:

$$OBT 2(CAB+BTO)$$

1. $$\measuredangle ABP=\measuredangle OPB+\measuredangle OTB=\measuredangle OPB+\measuredangle TOB=$$ $$=\measuredangle OPB+\measuredangle OPB+\measuredangle OBP=3\measuredangle OBP.$$
2. $$\measuredangle POA=\measuredangle OTB+\measuredangle OAB=\measuredangle OTB+\measuredangle OBA=$$ $$=\measuredangle OTB+\measuredangle OTB+\measuredangle BOT=3\measuredangle BOT.$$
3. $$\measuredangle OBT=\measuredangle OAB+\measuredangle AOB=\measuredangle ABO+2\measuredangle CAB=$$ $$=\measuredangle BTO+\measuredangle BOT+2\measuredangle CAB=2(\measuredangle BTO+\measuredangle CAB).$$